For combination/permutation type problems, think about the question logically.

We're dealing with committees, so this is a really a question about groups--we don't care about how the employees are arranged within the group.

The committee needs to contain 2 of the 5 female employees, as well as the male employee.

Since the order of the employees doesn't matter, let's represent the committee as: M F F

With the one spot for the male employee taken up, we have two remaining slots for 5 female employees.

The first slot can be filled by any of the 5 female employees, leaving us with 4 for the second slot. That's 5*4 = 20

BUT

Since order doesn't matter with group problems, we don't want any duplicate committees. A committee with Julie in the first slot and Michelle in the second is the same as one with Michelle in the first and Julie in the second. So we need to divide 2 (technically 2!) to get rid of the redundant committees.

That leaves \(\frac{20}{2}=10\) groups of female employees. Multiply by the lone male employee, and that's 10*1 = 10. Answer: A

If you'd prefer to use combinations

If you're more comfortable with the combinations/permutations approach, you'll take the number of groups of two women of the five female employees: (5C2), and multiply by the "group" of the lone male employee (1),

5C2 = \(\frac{5!}{3!2!}*1 = 10\)

Answer: A